Cauchy's Integral Formula implies Holomorphicity? Is the converse direction of the Cauchy Integral Formula true? Meaning,
if $f:\mathbb{C}\supseteq U\rightarrow\mathbb{C}$, and
$$\forall a \in U \space \space f(a) = \frac{1}{2\pi i}\int_{|z-a|=R}\frac{f(z)}{z-a}dz$$
does it mean that f is complex-differenatiable?
 A: If you parametrize your integral, you see that your equation is equivalent to
$$ f(a) = \dfrac{1}{2\pi} \int_0^{2\pi} f(a + r e^{i\theta})\; d\theta$$
and this is true for harmonic functions, not just holomorphic ones.
A: No the answer will be only harmonic, as example $f(z)=Re(z)$ the real part of $z$, we know using Cauchy-Riemann equation that a real valued holomorphic function will be constant.
but for all $a=\alpha+i\beta$ 
we have 
$$
\alpha=f(a)=\frac{1}{2\pi}\int_{0}^{2\pi} (\alpha+R\cos(\theta))d\theta=\frac{1}{2\pi}\int_{0}^{2\pi} f(a+Re^{i\theta})d\theta=\frac{1}{2\pi}\int_{|z-a|=R}\frac{f(z)}{z-a}dz
$$
but we have another theorem : morera theorem  witch say that if a continuous function $f$ on a connected open set $\Omega$ that verify :
$$
\oint_{\gamma} f(z)dz=0 \qquad \forall \gamma \, \textrm{  closed piecewise $C^1$ curve}
$$
then $f$ will be holomorphic.
A: Well...yes, because then you would have an expression for the nth derivative of f, making f infinitely differentiable/analytic in U and thus holomorphic.
https://en.wikipedia.org/wiki/Analytic_function
